NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2

In this page we have NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2. This exercise has question about how to find the squares of the numbers, how to find Pythagorean triplet.Hope you like them and do not forget to like , social share
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NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2

Question 1
Find the square of the following numbers.
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46 Answer
(i) 32^{2}
We can find the square using direct multiplication
= 32 x 32 = 1024
But above method can be cumbersome to calculate. We can calculate such values in the another better way f
Since, 32 can be written as (30+2)
So, 32^{2} = (30+2)^{2} = (30+2)(30+2)
Now we know the identity
(a+b)^{2}= a^{2} + b^{2} +2ab
= 30^{2} + 2 x 30 x 2 + 2^{2}
= 900 + 120 + 4 = 1024
(ii) (35)^{2 }= (30+5)^{2}
Solving on similar lines as above
= 30^{2} + 2 x 30 x 5 + 5^{2}
= 900 + 300 + 25 = 1225
(iii) 86^{2} = (80 + 6)^{2}
= 80^{2} + 2 x 80 x 6 + 6^{2}
= 6400 + 960 + 36 = 7396
(iv) 93^{2 }= (90+3)^{2}
= 90 ^{2} +2 x 90 x 3 + 3 ^{2}
= 8100 + 540 + 9 = 8649
(v) 71 ^{2} = (70 + 1) ^{2}
= 70^{2} +2 x 70 x 1 + 1 x 1
= 4900 + 140 + 1 = 5040 + 1 = 5041
(vi) 46^{2 }= (40+6)^{2}
= 40 ^{2} + 2 x 40 x 6 + 6^{2}
= 1600 + 480 + 36 = 2080 + 36 = 2116

Question 2
Write a Pythagorean triplet whose one member is:
(i) 6
(ii) 14
(iii) 16
(iv) 18 Answer
As we know 2n, n ^{2} + 1 and n^{2} - 1 form a Pythagorean triplet for any number, n > 1.
(i) If we take n ^{2} + 1 or n^{2} - 1 to be 6 then then the value of n will not integer(n^{2 } will be 5 or 7)
So we can 2n = 6
Therefore, n = 3
And, n^{2} + 1 = 3^{ 2 }+ 1= 9 + 1 = 10
And,n^{ 2 }- 1 = 3^{ 2 }- 1 = 9 - 1 = 8
Test: 6^{ 2 }+ 8^{ 2 }= 36 + 64 = 100 = 10^{2}
Hence, the triplet is 6, 8, and 10 Answer
(ii) If we take n ^{2} + 1 or n^{2} - 1 to be 14 then then the value of n will not integer(n^{2} will be 15 or 13)
So we can take 2n= 14, therefore, n = 7
Now, n ^{2 }+ 1 = 7 ^{2 }+ 1 = 49 + 1 = 50
And, n^{2 }- 1 = 7 ^{2 }- 1 = 49 - 1 = 48
Test: 14 ^{2 }+ 48 ^{2 }= 196 + 1304 = 2500 = 50 ^{2 }
Hence, the triplet is 14, 48, and 50 Answer
(iii) If we take n ^{2} + 1 or n^{2} - 1 to be 16 then then the value of n will not integer(n^{2} will be 17 or 15)
Let us assume 2n = 16, then n = 8
Now, n^{2 }+ 1 = 8^{ 2 }+ 1 = 64 + 1 = 65
And, n ^{2 }- 1 = 8^{ 2 }- 1 = 64 - 1 = 63
Test: 16^{2 }+ 63^{ 2 }= 256 + 3969 = 4225 = 65 ^{2 }
Hence, the triplet is 16, 63, and 65 Answer
(iv) If we take n ^{2} + 1 or n^{2} - 1 to be 18 then then the value of n will not integer(n^{2} will be 19 or 17)
Let us assume 2n = 18, therefore, n = 9
Now, n ^{2 }+ 1 = 9 ^{2 }+ 1 = 81 + 1 = 82
And, n ^{2 }- 1 = 9 ^{2 }- 1 = 81 - 1 = 80
Test: 18 ^{2 }+ 80 ^{2 }= 324 + 6400 = 6724 = 82 ^{2 }

Summary

NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.2 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also Download this assignment as pdf

This chapter 6 has total 4 Exercise 6.1 ,6.2,6.3 and 6.4. This is the Second exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below